Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $12.6$ years; the standard deviation is $2.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $6.8$ and $9.7$ years.
$12.6$ $9.7$ $15.5$ $6.8$ $18.4$ $3.9$ $21.3$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $12.6$ years. We know the standard deviation is $2.9$ years, so one standard deviation below the mean is $9.7$ years and one standard deviation above the mean is $15.5$ years. Two standard deviations below the mean is $6.8$ years and two standard deviations above the mean is $18.4$ years. Three standard deviations below the mean is $3.9$ years and three standard deviations above the mean is $21.3$ years. We are interested in the probability of a lion living between $6.8$ and $9.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the mean. The probability of a particular lion living between $6.8$ and $9.7$ years is $\color{orange}{13.5\%}$.